Question

You are given the following standard enthalpies of formation at 25°C. HF(aq) \text{ } -320.1 \text{ kJ/mol} OH$^-$(aq) \text{ } -229.6 \text{ kJ/mol} F$^-$(aq) \text{ } -329.1 \text{ kJ/mol} H$_2$O(l) \text{ } -285.8 \text{ kJ/mol} Calculate the standard enthalpy of reaction ($\Delta H^\circ_{rxn}$, in kJ to one decimal place) for the following: HF(aq) + OH$^-$(aq) $\rightarrow$ F$^-$(aq) + H$_2$O(l) Whether it is positive or negative matters. Do not enter units.

          You are given the following standard enthalpies of formation at 25°C.
HF(aq)  \text{ } -320.1 \text{ kJ/mol}
OH$^-$(aq) \text{ } -229.6 \text{ kJ/mol}
F$^-$(aq) \text{ } -329.1 \text{ kJ/mol}
H$_2$O(l) \text{ } -285.8 \text{ kJ/mol}
Calculate the standard enthalpy of reaction ($\Delta H^\circ_{rxn}$, in kJ to one decimal place) for the
following:
HF(aq) + OH$^-$(aq) $\rightarrow$ F$^-$(aq) + H$_2$O(l)
Whether it is positive or negative matters. Do not enter units.

You are given the following standard enthalpies of formation at 25°C.
HF(aq) -320.1 kJ/mol
OH^-(aq) -229.6 kJ/mol
F^-(aq) -329.1 kJ/mol
H2O(l) -285.8 kJ/mol
Calculate the standard enthalpy of reaction (Δ H^∘rxn, in kJ to one decimal place) for the
following:
HF(aq) + OH^-(aq) → F^-(aq) + H2O(l)
Whether it is positive or negative matters. Do not enter units.

Added by Brian J.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

07/02/2023

Step 1

Step 1: Write the balanced chemical equation for the reaction: HF(aq) + OH^(-)(aq) -> F^(-)(aq) + H2O(l) Show more…

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You are given the following standard enthalpies of formation at 25deg C. HF(aq),-320.1k(J)/(m)ol OH^(-)(aq),-229.6k(J)/(m)ol F^(-)(aq),-329.1k(J)/(m)ol H_(2)O(l),-285.8k(J)/(m)ol Calculate the standard enthalpy of reaction ( DHdeg _(rxn), in kJ to one decimal place) for the following: HF(aq)+OH^(-)(aq)->F^(-)(aq)+H_(2)O(l) Whether it is positive or negative matters. Do not enter units. You are given the following standard enthalpies of formation at 25'C. HF(aq) 320.1kJ/mol OH(aq) 229.6kJ/mol F-aq 329.1 kJ/mol HO(l) 285.8 kJ/mol Calculate the standard enthalpy of reaction (DH'rxn, in kJ to one decimal place) for the following: HFaq+OH-aq>Faq+HOl Whether it is positive or negative matters. Do not enter units.